Linear Algebra Exercises-n-Answers.pdf

Chapter Two: Vector Spaces
Subsection Two.I.1: Definition and Examples
Two.I.1.17 ( (a) 0 ) + 0x + 0x 2 + 0x 3
0 0 0 0
(b)
0 0 0 0
(c) The constant function f(x) = 0
(d) The constant function f(n) = 0
Two.I.1.18 (a) 3 + 2x − x 2 (b)
( )
−1 +1
0 −3
(c) −3e x + 2e −x
Two.I.1.19 Most of the conditions are easy to check; use Example 1.3 as a guide. Here are some
comments.
(a) This is just like Example 1.3; the zero element is 0 + 0x.
(b) The zero element of this space is the 2×2 matrix of zeroes.
(c) The zero element is the vector of zeroes.
(d) Closure of addition involves noting that the sum
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
x 1 x 2 x 1 + x 2
⎜y 1
⎟
⎝z 1
⎠ + ⎜y 2
⎟
⎝ z 2
⎠ = ⎜ y 1 + y 2
⎟
⎝ z 1 + z 2
⎠
w 1 w 2 w 1 + w 2
is in L because (x 1 +x 2 )+(y 1 +y 2 )−(z 1 +z 2 )+(w 1 +w 2 ) = (x 1 +y 1 −z 1 +w 1 )+(x 2 +y 2 −z 2 +w 2 ) = 0+0.
Closure of scalar multiplication is similar. Note that the zero element, the vector of zeroes, is in L.
Two.I.1.20 In each item the set is called Q. For some items, there are other correct ways to show that
Q is not a vector space.
⎞
⎛
⎞
⎛
⎞
(a) It is not closed under addition. ⎛
⎝ 1 0
0
⎠ ,
⎝ 0 1⎠ ∈ Q
0
⎝ 1 1⎠ ∉ Q
0
(b) It is not closed under addition. ⎛
⎝ 1 0
⎞
⎛
⎞
⎛
⎞
(c) It is not closed under addition.
( ) 0 1
,
0 0
⎠ , ⎝ 0 1⎠ ∈ Q
0 0
( ) 1 1
∈ Q
0 0
(d) It is not closed under scalar multiplication.
(e) It is empty.
⎝ 1 1⎠ ∉ Q
0
( ) 1 2
∉ Q
0 0
1 + 1x + 1x 2 ∈ Q − 1 · (1 + 1x + 1x 2 ) ∉ Q
Two.I.1.21 The usual operations (v 0 + v 1 i) + (w 0 + w 1 i) = (v 0 + w 0 ) + (v 1 + w 1 )i and r(v 0 + v 1 i) =
(rv 0 ) + (rv 1 )i suffice. The check is easy.
Two.I.1.22
No, it is not closed under scalar multiplication since, e.g., π · (1) is not a rational number.
Two.I.1.23 The natural operations are (v 1 x + v 2 y + v 3 z) + (w 1 x + w 2 y + w 3 z) = (v 1 + w 1 )x + (v 2 +
w 2 )y + (v 3 + w 3 )z and r · (v 1 x + v 2 y + v 3 z) = (rv 1 )x + (rv 2 )y + (rv 3 )z. The check that this is a vector
space is easy; use Example 1.3 as a guide.
Two.I.1.24 The ‘+’ operation is not commutative; producing two members of the set witnessing this
assertion is easy.